洛谷 P3711 仓鼠的数学题 [伯努利数 fft]

P3711 仓鼠的数学题

题意：

\[S_m(x) = \sum_{k=0}^x k^m, 0^0=1\quad 求 \sum_{m=0}^n S_m(x)a_m \]

的答案多项式\(\sum_{i=0}^{n+1}c_ix^i\)各项系数

---

一开始用了\(B^-\)，然后后面要展开\((x+1)^k\)，完全不会做

和出题人fjzzq2002讨论了一下，原来标程用的是\(B^+\)，不需要展开了

那就很简单了...不想写过程了，最后的结果就是

\[C_t = \frac{1}{t!} \sum_{m=0}^{n+t} f_m h_{m+1-t} \\ f(x) = \sum_{i=0}^n a_i i!,\ g(x) = \sum_{i=0}^n \frac{B_i^+}{i!},\ h_i = g_{n+1-i} \]

然后我发现用\(B^+\)好像没有常数项啊？但是用\(B^-\)写暴力计算确实有常数项

然后发现常数项就是\(a_0\)，交上就过了...

update:去问了张队长，然后意识到，\(B^+\)算的和是从1开始...然后常数项就是\(a_0 0^0 = a_0\)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int N = (1<<20) + 5, P = 998244353, mo = P, inv2 = (P+1)/2;
const double PI = acos(-1.0);
inline int read(){
    char c=getchar(); int x=0,f=1;
    while(c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();}
    return x*f;
}

inline int Pow(ll a, int b) {
	ll ans = 1;
	for(; b; b>>=1, a=a*a%P)
		if(b&1) ans=ans*a%P;
	return ans;
}
namespace fft {
	int rev[N], g = 3;
	void dft(int *a, int n, int flag) {
		int k = 0; while((1<<k) < n) k++;
		for(int i=0; i<n; i++) {
			rev[i] = (rev[i>>1]>>1) | ((i&1)<<(k-1));
			if(i < rev[i]) swap(a[i], a[rev[i]]);
		}
		for(int l=2; l<=n; l<<=1) {
			int m = l>>1, wn = Pow(g, flag==1 ? (P-1)/l : P-1-(P-1)/l);
			for(int *p = a; p != a+n; p += l)
				for(int k=0, w=1; k<m; k++, w = (ll)w*wn %P) {
					int t = (ll) p[k+m] * w %P;
					p[k+m] = (p[k] - t + P) %P;
					p[k] = (p[k] + t) %P;
				}
		}
		if(flag == -1) {
			ll inv = Pow(n, P-2);
			for(int i=0; i<n; i++) a[i] = a[i] * inv %P;
		}
	}
	
	int t[N];
	void inverse(int *a, int *b, int l) {
		if(l == 1) {b[0] = Pow(a[0], P-2); return;}
		inverse(a, b, l>>1);
		int n = l<<1;
		for(int i=0; i<l; i++) t[i] = a[i], t[i+l] = 0;
		dft(t, n, 1); dft(b, n, 1);
		for(int i=0; i<n; i++) b[i] = (ll) b[i] * (2 - (ll) b[i] * t[i] %P + P) %P;
		dft(b, n, -1); for(int i=l; i<n; i++) b[i] = 0;
	}

	void mul(int *a, int *b, int n) {
		dft(a, n, 1); dft(b, n, 1);
		for(int i=0; i<n; i++) a[i] = (ll) a[i] * b[i] %P;
		dft(a, n, -1);
	}
}

int n, a[N], len;
ll inv[N], fac[N], facInv[N];
int q[N], b[N], f[N], g[N];
int main() {
	freopen("in", "r", stdin);
	n=read();
	for(int i=0; i<=n; i++) a[i] = read();

	inv[1] = fac[0] = facInv[0] = 1;
	for(int i=1; i<=n+2; i++) {
		if(i != 1) inv[i] = (P - P/i) * inv[P%i] %P;
		fac[i] = fac[i-1] * i %P;
		facInv[i] = facInv[i-1] * inv[i] %P;
	}

	len = 1; while(len <= n+1) len <<= 1;
	for(int i=0; i<=n+1; i++) q[i] = facInv[i+1];
	fft::inverse(q, b, len); 
	for(int i=0; i<=n+1; i++) b[i] = b[i] * fac[i] %P;
	b[1] = inv2;


	for(int i=0; i<=n; i++) f[i] = (ll) a[i] * fac[i] %P;
	for(int i=0; i<=n; i++) g[n+1-i] = (ll) b[i] * facInv[i] %P;
	while(len <= n+n+1) len <<= 1;
	fft::mul(f, g, len);
	printf("%d ", a[0]);
	for(int i=1; i<=n+1; i++) printf("%lld ", f[n+i] * facInv[i] %P);
}